题意：

给你三个数组 $a, b, c$ ，让你从每个数组中选择一个数 $x, y, z$ ，使得 $x-y)^2+(x-z)^2+(y-z)^2$ 最小，求这个最小值。

思路：

由于答案一定是 $x≤y≤zx\le y\le z$ 的形式(当然这里的 $x, y, z$ 与题意的 $x, y, z$ 不同)，所以我们就可以跑一个 $3!$ 排列，让他们分别为 $x, y, z$ ，让后枚举其中一个二分另外俩就行了。
具体实现可以写个函数就比较方便了。
我这里写的是枚举 $x$ ，让后分两种即找 $y$ 和找 $z$ ，如果找 $y$ 的话，假设现在已经有 $x≤yx\le y$ 了，那么找的 $z$ 根据平方的性质，最好是 $x + (y - x) / 2$ 附近的位置，二分找即可，对于 $z$ 同理。
~~总结：这个题瞎搞都能过。~~

// Problem: D. Xenia and Colorful Gems
// Contest: Codeforces - Codeforces Round #635 (Div. 2)
// URL: https://codeforces.com/contest/1337/problem/D
// Memory Limit: 256 MB
// Time Limit: 3000 ms
// 
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const LL inf=0x3f3f3f3f3f3f3f3f;
const double eps=1e-6;int n1,n2,n3;
LL a[N],b[N],c[N];LL get(int i,int j,int k) {//cout<<i<<' '<<j<<' '<<k<<endl;return (a[i]-b[j])*(a[i]-b[j])+(a[i]-c[k])*(a[i]-c[k])+(b[j]-c[k])*(b[j]-c[k]);
}LL solve() {LL ans=inf;for(int i=1;i<=n1;i++) {int val1=a[i];int pos1=lower_bound(b+1,b+1+n2,val1)-b;if(pos1<=n2) {int pos2=lower_bound(c+1,c+1+n3,a[i]+(b[pos1]-a[i])/2)-c;if(pos2<=n3) ans=min(ans,get(i,pos1,pos2));			pos2--;if(pos2>=1) ans=min(ans,get(i,pos1,pos2));}pos1++;if(pos1<=n2) {int pos2=lower_bound(c+1,c+1+n3,a[i]+(b[pos1]-a[i])/2)-c;if(pos2<=n3) ans=min(ans,get(i,pos1,pos2));			pos2--;if(pos2>=1) ans=min(ans,get(i,pos1,pos2));}pos1-=2;if(pos1>=1) {int pos2=lower_bound(c+1,c+1+n3,b[pos1]+(a[i]-b[pos1])/2)-c;if(pos2<=n3) ans=min(ans,get(i,pos1,pos2));			pos2--;if(pos2>=1) ans=min(ans,get(i,pos1,pos2));}pos1=lower_bound(c+1,c+1+n3,val1)-c;if(pos1<=n3) {int pos2=lower_bound(b+1,b+1+n2,a[i]+(c[pos1]-a[i])/2)-b;if(pos2<=n2) ans=min(ans,get(i,pos2,pos1));			pos2--;if(pos2>=1) ans=min(ans,get(i,pos2,pos1));}pos1++;if(pos1<=n3) {int pos2=lower_bound(b+1,b+1+n2,a[i]+(c[pos1]-a[i])/2)-b;if(pos2<=n2) ans=min(ans,get(i,pos2,pos1));			pos2--;if(pos2>=1) ans=min(ans,get(i,pos2,pos1));}pos1-=2;//cout<<pos1<<' '<<c[pos1]<<' '<<a[i]<<endl;if(pos1>=1) {int pos2=lower_bound(b+1,b+1+n2,c[pos1]+(a[i]-c[pos1])/2)-b;if(pos2<=n2) ans=min(ans,get(i,pos2,pos1));			pos2--;if(pos2>=1) ans=min(ans,get(i,pos2,pos1));}}return ans;
}int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);int _; scanf("%d",&_);while(_--) {scanf("%d%d%d",&n1,&n2,&n3);for(int i=1;i<=n1;i++) scanf("%lld",&a[i]);for(int i=1;i<=n2;i++) scanf("%lld",&b[i]);for(int i=1;i<=n3;i++) scanf("%lld",&c[i]);sort(a+1,a+1+n1); sort(b+1,b+1+n2); sort(c+1,c+1+n3);n1=unique(a+1,a+1+n1)-a-1;n2=unique(b+1,b+1+n2)-b-1;n3=unique(c+1,c+1+n3)-c-1;printf("%lld\n",solve());}return 0;
}
/**/